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Cap 21 Fisica Solucionario

Apr 07, 2018

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    21.1:

    C103.20chargeandg00.8 9lead

    v!!m

    a) .100.2C106.1

    C1020.3 1019

    9

    e v!v

    v!

    n

    b) 1058.8and1033.2207

    g00.8 13

    lead

    e22lead

    v!v!v!nnNn A

    21.2: s10s100ands000,20current4!!! Qt

    Q = It= 2.00 C

    .1025.1C1060.1

    19

    19ev!

    v!

    Qn

    21.3: The mass is primarily protons and neutrons of271067.1 v!m kg, so:

    28

    27nandp 1019.4kg101.67

    kg70.0

    v!v! n

    About one-half are protons, so e28

    p 1010.2 nn !v! and the charge on the electrons is

    given by: .1035.3)1010.2()1060.1(92819 v!vvv! Q

    21.4: Mass of gold = 17.7 g and the atomic weight of gold is 197 mol.g So the number

    of atoms .1041.5)1002.6(mol 22molg197

    g7.1723 v!vv!vA

    N

    a)2422

    p 1027.41041.579 v!vv!n

    C106.83C1060.1 519p v!vv!nq

    b) .1027.424

    pe v!! nn

    21.5: .electrons101.08atomsH106.021.80mol80.12423 v!vv!

    C.101.73C1060.1101.08charge 51924 v!vvv!

    21.6: First find the total charge on the spheres:

    1043.1)2.0)(1057.4(444

    1 162210

    2

    02

    2

    0

    v!v!!! Frqr

    q

    F

    And therefore, the total number of electrons required is

    890.C101.60C1043.1 1916 !vv!! eqn

    21.7: a) Using Coulombs Law for equal charges, we find:

    .C1042.7C105.5m)150.0(4

    1N220.0

    7213

    2

    2

    0

    v!v!!! qq

    b) When one charge is four times the other, we have:

  • 8/4/2019 Cap 21 Fisica Solucionario

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    1071.310375.1m)150.0(

    4

    4

    1220.0

    7213

    2

    2

    0

    v!v!!! qq

    F

    So one charge is71071.3 v C, and the other is C.10484.1 6v

    21.8:a) The total number of electrons on each sphere equals the number of protons.

    .1025.7molkg026982.0

    kg0250.013 24pe v!vv!! ANnn

    b) For a force of41000.1 v N to act between the spheres,

    .1043.8m)08.0(N)10(44

    1N10

    424

    02

    2

    0

    4 v!!!! qr

    q

    F

    15

    e 1027.5 v!!d eqn

    c)10

    e 107.27isvdn of the total number.

    21.9: The force of gravity must equal the electric force.

    .m08.5m8.25)sm8.9(kg)1011.9(

    C)1060.1(

    4

    1

    4

    1 231

    219

    0

    2

    2

    2

    0

    !!v

    v!!

    r

    rr

    q

    mg

    21.10: a) Rubbing the glass rod removes electrons from it, since it becomes positive.

    kg.1027.4)electronkg1011.9(electrons)104.69(

    electrons1069.4)Celectrons1025.6(C)107.50(nC50.7

    203110

    10189

    v!vv

    v!vv!

    The rods mass decreases by kg.1027.420v

    b) The number of electrons transferred is the same, but they are addedto the mass of the

    plastic rod, which increases by kg.1027.420v

    21.11: positive.isanddirection-theinbemustsodirection,-theinis 112 qxx FFTT

    n750.00400.00200.0

    ,

    2

    2

    1

    2

    23

    32

    2

    13

    31

    21

    !!

    !!

    qq

    r

    qqk

    r

    qqkFF

    21.12: a)2

    2

    6

    0

    2

    21

    0 m)30.0(

    C)10550.0(

    4

    1N200.0

    4

    1 q

    r

    qq

    v!!

    C.1064.3 62 v! q b) 200.0!F N, and is attractive.

    21.13: Since the charges are equal in sign the force is repulsive and of magnitude:

    N172.0m)800.0(4

    C)1050.3(2

    0

    26

    2

    2

    !v

    !!

    r

    kqF

  • 8/4/2019 Cap 21 Fisica Solucionario

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    21.14: We only need the y-components, and each charge contributes equally.

    0.6).sinsince(N173.0sinm)500.0(

    C)104(C)100.2(

    4

    12

    66

    0

    !!vv

    !

    F

    Therefore, the total force is N35.02 !F , downward.

    21.15: theinbothareand 32 FF TT x-direction.

    N10124.1,N10749.64

    2

    13

    31

    3

    5

    2

    12

    21

    2

    v!!v!!r

    qqkF

    r

    qqkF

    v!! thein,108.1 432 FFF x-direction.

    21.16:

    N100.0m60.0

    C)100.2(C)10.20()CmN109(2

    66229

    21 !vvv

    !

    F

    1QF is equal and opposite to QF1 (Ex. 21.4), so N17.0

    N23.0

    1

    1

    !

    !

    yQ

    xQ

    F

    F

    Overall:

    N27.0N17.0N100.0

    N23.0

    !!

    !

    y

    x

    F

    F

    The magnitude of the total force is N.35.0N27.0N23.0 22 ! The direction of theforce, as measured from the +y axis is

    Q

    4027.0

    23.0

    tan1

    !!

    21.17: 2FT

    is in the direction. x

    N10.37N37.3N00.7

    N00.7and

    N37.3so,N37.3

    23

    32

    22

    12

    21

    2

    !!!

    !!

    !!!

    xxx

    xxxx

    x

    FFF

    FFFF

    Fr

    qqkF

    For xF3 to be negative, 3q must be on the x-axis.

    m144.0som,144.0so,3

    31

    2

    31

    3 !!!! xF

    qqkxx

    qqkF

    21.18: The charge 3q must be to the right of the origin; otherwise both 32 and qq would

    exert forces in the + x direction. Calculating the magnitude of the two forces:

  • 8/4/2019 Cap 21 Fisica Solucionario

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    direction.in the375.3

    m)200.0(

    C)1000.5C)(1000.3)(Cm109(

    4

    12

    66229

    2

    12

    21

    0

    21

    x

    r

    qq

    F

    !

    vvv!!

    xr

    rF

    !

    vvv

    !

    We need

    ! FF

    !

    r

    r gfr g.

    8...

    .

    !

    !

    !

    r

    r

    21.19:

    FFF !! FFFTTT

    since they are acting in the same direction at

    m!y so,

    o

    !

    v!

    v

    vv!

    "

    21.20: #$

    #

    $

    FFF !! FFFTTT

    since they are acting in opposite directions at

    x= 0 so,

    righttheto0000

    000

    000

    000000

    %

    &

    '

    &

    '

    '

    0

    v!

    vvv!

    (

    21.21: a)

    b) )0))1

    ))

    1 )) xa

    qQa

    xa

    qQ

    22

    yx

    !!

    c) c34

    ya

    q 5

    6

    7

    8

    x y !!

  • 8/4/2019 Cap 21 Fisica Solucionario

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    d)

    21.22: a)

    b)))

    9

    /@99

    A

    99

    A

    !

    !

    ! yx F

    xa

    qQx

    xa

    qQ

    F

    c) At x = 0, F= 0.

    d)

    21.23:

    b) thebelow45ofangleanat24

    1221

    4

    12

    24

    12

    2

    02

    2

    02

    2

    0

    r!!L

    q

    L

    q

    L

    q

    F

  • 8/4/2019 Cap 21 Fisica Solucionario

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    positive x-axis

    21.24: a) particle.thetowarddown,CN432m)250.0(

    C)1000.3(

    4

    1

    4

    12

    9

    0

    2

    0

    !v

    !

    r

    q

    E

    b) m.50.1)CN0.12(C)1000.3(

    41

    41CN00.12

    9

    0

    2

    0

    !v!!!

    r

    rq

    E

    21.25: Let +x-direction be to the right. Find :xa

    ? A CN5.23C)10602.1(2N)10516.7(le t.thetoissopositive,ischarge,direction-le tthetois

    N10516.7

    sm101.132gives

    ?,s1065.2,sm101.50,sm1050.1

    1918

    18

    29

    0

    633

    0

    !vv!!

    v!!

    v!!

    !v!v!v!

    qFB

    x

    C aF

    atavv

    atvv

    xx

    xxxx

    xxx

    EFTT

    21.26: (a)2

    21 atx !

    CN69.5

    C106.1

    )sm101.00(kg)1011.9(

    sm1000.1s)103.00(

    m)50.4(22

    19

    21231

    212

    26-2

    !

    v

    vv!!!

    v!v

    !!

    q

    ma

    q

    FE

    t

    xa

    The force is up, so the electric field must be downwardsince the electron is negative.

    (b) The electrons acceleration is ~

    11

    10 g, so gravity must be negligibly smallcompared to the electrical force.

    21.27: a) negative.issign,C1019.2CN650

    )sm8.9(kg)00145.0( 52

    v!!! qmgEq

    b) C,N1002.1C101.60

    )sm9.8(kg)1067.1( 719

    227

    v!v

    v!! EmgqE

    21.28: a) .CN1004.1m)1000.6(

    C)1060.126(

    4

    1

    4

    1 11210

    19

    0

    2

    0

    v!v

    vv!!

    r

    q

    E

    b) CN1015.5m)1029.5(

    C)1060.1(

    4

    1

    4

    1 11211

    19

    0

    2

    0

    proton v!vv

    !!

    r

    q

    E

    21.29: a) Fq andC,100.556v! is downward with magnitude

    C,N1013.1There ore,N.1020.6 49 v!!v qFD

    upward.

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    b) If a copper nucleus is placed at that point, it feels an upward force of magnitude

    N.105.24CN101.13C106.129 22419 v!vv!! qEF

    21.30: a) The electric field of the Earth points toward the ground, so a NEGATIVE charge

    will hover above the surface.

    C.92.3CN150

    )sm8.9(kg)0.60( 2 !!! qqEmg

    b) N1038.1m)00.100(

    C)92.3(

    4

    1

    4

    1 72

    2

    0

    2

    2

    0

    v!!!r

    q

    F The magnitude of the charge is too

    great for practical use.

    21.31: a) Passing between the charged plates the electron feels a force upward, and justmisses the top plate. The distance it travels in the y-direction is 0.005 m. Time of flight

    s1025.1 8sm1060.1

    m0200.06

    vv!!! t and initialy-velocity is zero. Now,

    .sm1040.6s)1025.1(m005.0so21328

    212

    210 v!v!! aaattvy y But also.CN364

    C1060.1

    )sm1040.6kg)(1011.9(19

    21331

    !!!!

    v

    vvEa

    emeE

    mF

    b) Since the proton is more massive, it will accelerate less, and NOT hit the plates. Tofind the vertical displacement when it exits the plates, we use the kinematic equations

    again:

    m.1073.2s)1025.1(2

    1

    2

    1 628

    p

    2 v!v!!m

    eEaty

    c) As mention in b), the proton will not hit one of the plates because although the

    electric force felt by the proton is the same as the electron felt, a smaller acceleration resultsfor the more massive proton.

    d) The acceleration produced by the electric force is much greater thang; it isreasonable to ignore gravity.

    21.32: a)

    jjE )CN10813.2(

    m0400.0

    C)1000.5()CmN109(

    4

    4

    2

    9229

    2

    10

    1

    1v!

    vv!!

    r

    qT

    CN1008.1

    m)0400.0(0.0300m

    C)1000.3()CmN109( 4

    22

    9229

    2

    2

    2

    2v!

    vv!!

    r

    qET

    The angle of ,2ET

    measured from the axis,-x is r! 9.126tan180cm3.00

    cm00.41 Thus

    ji

    jiE

    )CN10(8.64)CN106.485(

    )9.126sin9.126cos()CN10080.1(

    33

    4

    2

    vv!

    rrv!T

    b) The resultant field is

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    ji

    jiEE

    )CN1095.1()CN10485.6(

    )CN1064.8CN10813.2()CN10485.6(

    43

    343

    21

    vv!

    vvv!TT

    21.33: Let x be to the right and y be downward.

    Use the horizontal motion to find the time when the electron emerges from the field:

    sm1079.1

    sm1000.8gives2

    ,s1025.1,0m,0.0050

    sm1060.1

    s1025.1gives

    ,sm1060.1,0,m0200.0

    622

    50

    0

    8

    0y0

    6

    82

    21

    00

    6

    00

    v!!

    v!

    !

    !v!!!

    v!

    v!!

    !v!!!

    yx

    y

    yy

    y

    x

    xx

    xx

    vvv

    vtvv

    yy

    vtvyy

    v

    ttatvxx

    tvaxx

    21.34: a) .CN8.17)14()11(so,CN14CN11 22 !!! EjiET

    r!r!! 128so,8.51)1114(tan 1 counterclockwise from the x-axis

    b) )repulsive(52at)iN,1045.4)C105.2()CN8.17(so89 rv!v!! FqEF

    TT

    ).repulsive(128at)ii r

    21.35: a) !!v!v!! eEFgmF e30231

    eg .N1093.8)sm(9.8kg)1011.9(

    .N1060.1)CN1000.1()C1060.1( 15419 v!vv Yes, ok to neglect gF because .ge FF ""

    b) kg1063.1N106.1CN10

    1615

    e

    4

    v!!v!! mmgFF

    .101.79 e

    14mm v! c) No. The field is uniform.

    21.36: a) .CN148)1050.1()C1060.1(

    )kg1067.1()m0160.0(2

    2

    1

    2

    12619

    2722 !

    vv

    v!!!

    sEt

    m

    eEatx

    p

    b) .sm1013.24

    0 v!!! tm

    eEatvv

    p

    21.37: a) jr ,20

    35.1tan 1 !!

    T b) jir

    22

    22,

    42.12tan 1 !!

    c) jir 92.039.0,112.9radians97.110.1

    6.2tan 1 !r!!

    (Second quadrant).

    21.38: a) .N1082.9,CN61417v!!! qEFE

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    b) .N103.2)100.1(48210

    0

    2 v!v! eF c) Part (b) >> Part (a), so the electron hardly notices the electric field. A person in the

    electric field should notice nothing if physiological effects are based solely on magnitude.

    21.39: a) Let x be east.ET

    is west and q is negative, so FT

    is east and the electron speeds up.

    sm1033.6gives)(2

    ?m,375.0,sm10638.2,sm1050.4

    sm10638.2)kg10109.9()N10403.2(

    N10403.2)m50.1()C10602.1(||

    5

    0

    2

    0

    2

    0

    2115

    0

    2113119

    1919

    v!!

    !!v!v!

    v!vv!!

    v!v!!

    xxxx

    xxx

    xx

    x

    vxxavv

    vxxav

    mFa

    EqF

    b) FT

    so0"q is west and the proton slows down.

    sm1059.1gives)(2

    ?,m375.0,sm10436.1,sm1090.1

    sm10436.1)kg10673.1()N10403.2(

    N10403.2)m50.1()C10602.1(||

    4

    0022

    0

    284

    0

    282719

    1919

    v!!

    !!v!v!

    v!vv!!

    v!v!!

    xxxx

    xxx

    xx

    x

    vxxavv

    vxxav

    mFa

    EqF

    21.40: Point charges1

    q (0.500 nC) and2

    q (8.00 nC) are separated by .m20.1!x The

    electric field is zero when !!!!

    2

    11

    2

    12)20.1(21)2.1(

    21

    2

    21

    1 rqrqEEr

    kq

    r

    kq

    072.02.17.5or0)2.1()2.1(2)(2.1)2.1(2 12

    11

    2

    11

    2

    1121

    2

    11

    2

    11 !! rrqrqrqqqrqrq24.04.0,24.0

    11!! rr is the point between.

    21.41:Two positive charges, q , are on the x-axis a distance a from the origin.

    a) Halfway between them, .0!E

    b) At any position

    "

    !

    axxa

    q

    xa

    q

    axxa

    q

    xa

    q

    axxa

    q

    xa

    q

    Ex

    ,)()(4

    1

    ,)()(4

    1

    ||,)()(4

    1

    ,

    22

    0

    22

    0

    22

    0

    For graph, see below.

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    21.42: The point where the two fields cancel each other will have to be closer to the

    negative charge, because it is smaller. Also, it cantt be between the two, since the twofields would then act in the same direction. We could use Coulombs law to calculate the

    actual values, but a simpler way is to note that the 8.00 nC charge is twice as large as the

    4.00 nC Charge. The zero point will therefore have to be a factor of farther from the

    8.00 nC charge for the two fields to have equal magnitude. Calling x the distance from the

    4.00 nC charge:

    m0.

    0.

    !

    !

    x

    xx

    21.43: a) Point chargeG

    q (2.00 nC) is at the origin and nC00.5(2 q ) is at !x

    i) At i t))

    H

    H

    H

    I !!!qkqk

    EP

    ii) At t))

    Q

    R

    Q

    Q

    !!!qkqk

    ES

    iii) At t))

    T

    T

    T

    U !!!qkqkV

    W

    b) !v!v!!

    FFeEF )iileft,N102.9N575106.1i) 1719

    right.N1048.6405106.1iii)right,N103.4N269106.1 17191719 vXvXvXv F

    21.44: A positive and negative charge, of equal magnitude q , are on the x-axis, a distanceY from the origin.

    a) Halfway between them, ,a

    b

    q

    E! to the left.

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    !! E atQ6.12

    4.128

    8.28tan 1 !

    ! up from x axis.

    21.47: a) At the origin,)

    )

    iiE !v

    !

    T

    b) At !! yx

    .C2133)m45.0(

    1

    )m15.0(

    1)C100.6(

    4

    122

    9

    0

    iiE !

    v!

    T

    c) At !! y

    v! jij

    5.0

    4.0

    )m5.0(

    15.0

    3.0

    )m5.0(

    1)m4.0(

    1)C100.6(

    4

    1222

    9

    0

    T

    !! EjiET

    and r clockwise from x

    axis.

    d) ).

    )..

    .

    .

    iE !v!!!!

    Eyx y

    T

    21.48: For a long straight wire,

    !v

    !!

    r

    rE

    21.49: a) For a wire of length a2 centered at the origin and lying along the y-axis, theelectric field is given by Eq. (21.10).

    iE

    1

    1

    0

    !

    axx

    T

    b) For an infinite line of charge:

    iE

    2

    0x

    !T

    Graphs of electric field versus position for both are shown below.

    21.50: For a ring of charge, the electric field is given by Eq. (21.8).

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    a) iE)4

    12322

    0 ax

    Qx

    !T

    so with

    iE !!!v!

    TxaQ

    b) ))

    j

    k

    l

    j m j

    iiEFF vn

    v

    n

    n

    n

    q

    21.51: For a uniformly charged disk, the electric field is given by Eq. (21.11):

    iE

    1

    11

    2 220

    o

    xR

    T

    The x -component of the electric field is shown below.

    21.52: The earths electric field is 150 , directly downward. So,

    ,10.00150

    0

    0

    v!!!!

    E and is negative.

    21.53: For an infinite plane sheet, E is constant and is given by

    E! directed

    perpendicular to the surface.

    c

    e.

    c.

    z!

    z

    !e

    so{

    |

    }

    ~

    !

    !

    directed toward the surface.

    21.54: By superposition we can add the electric fields from two parallel sheets of charge.

    a) .!E

    b) !E c)

    E !! directed downward.

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    21.55:

    21.56: The field appears like that of a point charge a long way from the disk and an infiniteplane close to the disks center. The field is symmetrical on the right and left (not shown).

    21.57: An infinite line of charge has a radial field in the plane through the wire, and

    constant in the plane of the wire, mirror-imaged about the wire:

    Cross section through the wire: Plane of the wire:

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    Length of vector does not depend on angle. Length of vector gets shorter at pointsfurther away from wire.

    21.58: a) Since field lines pass from positive charges and toward negative charges, we can

    deduce that the top charge is positive, middle is negative, and bottom is positive.

    b) The electric field is the smallest on the horizontal line through the middle charge, at

    two positions on either side where the field lines are least dense. Here the y-components ofthe field are cancelled between the positive charges and the negative charge cancels the x-component of the field from the two positive charges.

    21.59: a) ))

    !v!qdp , in the direction from and

    towards q .

    b) IfET

    is at r and the torque sinJpE! then:

    5.85636.9sin104.1(

    mN102.7

    sin 11

    9

    !rv

    v!!

    Jp

    21.60: a) 105.56)10(1.6m)109.(1119

    0 v!vv!! qpd

    b) ))

    v!vv!!

    pE

    Maximum torque:

    21.61: a) Changing the orientation of a dipole from parallel to perpendicular yields:

    !vv!rr!!( )CN10m)(1.6C100.5()0cos90cos(

    630pEpEUUU if

    J

    v

  • 8/4/2019 Cap 21 Fisica Solucionario

    16/32

    b) K.384.0)K103(1.38

    )108(2108

    2

    323

    2424 !

    vv

    !v!

    TkT

    21.62: 11.4

    m)100.3(2

    mC106.17m)1000.3(

    2

    )(39

    0

    309

    dipole3

    0

    dipole !

    vTI

    v!v!

    E

    x

    pxE

    .CN106v The electric force !vv!! )CN10C)(4.111060.1( 619qEF N106.58 13v and is toward the water molecule (negativex-direction).

    21.63: a)222222

    22

    22)4(

    2

    )4(

    )2()2(

    )2(

    1

    )2(

    1

    dy

    yd

    dy

    dydy

    dydy !

    !

    }

    !

    !

    3

    0

    222

    0

    222

    0 2)4(2)4(

    2

    4 y

    p

    dy

    y

    qd

    dy

    yd

    qEy

    b) This also gives the correct expression for yE since y appears in the full expressions

    denominator squared, so the signs carry through correctly.

    21.64: a) The magnitude of the field the due to each charge is

    ,))2(

    1

    44

    122

    0

    2

    0

    !!

    xd

    q

    r

    q

    E

    where dis the distance between the two charges. The x-components of the forces due to the

    two charges are equal and oppositely directed and so cancel each other. The two fields have

    equaly-components, so:

    xd

    qEE

    ysin

    )2(

    1

    4

    22

    22

    0

    !!

    where is the angle below the x-axis for both fields.

    ;)2(

    2sin

    22 xd

    d

    !

    thus

    2322

    02222

    0

    dipole))2((4)2(

    2

    )2(

    1

    4

    2

    xd

    qd

    xd

    d

    xd

    qE

    !

    !

    The field is the y directions.

    b) At large ,)2(,22

    dxx "" so the relationship reduces to the approximations

    3

    0

    dipole4 x

    qdE }

    21.65:

  • 8/4/2019 Cap 21 Fisica Solucionario

    17/32

    The dipoles attract.

    !! xxx FFF yyyy FFFF !!

    b)

    Opposite charges are closest so the dipoles attract.

    21.66: a)

    The torque is zero when pT

    is aligned either in the samedirection as ET

    or in the

    opposite directions

    b) The stable orientation is when p

    T

    is aligned in the same direction as E

    T

    c)

    21.67:

    r!! 48.6so00.250.1si

    Opposite charges attract and like charges repel.

    !! xxx FFF

    !!!

    !!

    v!vv

    !d

    !

    yyyy

    y

    FFFF

    FF

    k

    r

    qqkF

    (in the direction from the C-00.5 charge toward the -.5 charge).

    b)

  • 8/4/2019 Cap 21 Fisica Solucionario

    18/32

    The y components have zero moment arm and therefore zero torque.

    xF and xF2 both produce clockwise torques.

    .cos

    !! FFx

    N

    xF clockwise

    21.68: a) jiF sin41cos

    41 2

    13

    31

    0

    2

    13

    31

    0

    13 r

    r

    !

    T

    j

    i

    5

    3

    m)1016.0)((9.00

    nC)nC)(6.0000.5(

    4

    15

    4

    m)1016.0)((9.00

    nC)nC)(6.0000.5(

    4

    14

    0

    4

    0

    13 v

    v

    ! FT

    ji

    vv!

    T

    Similarly for the force from the other charge:

    jjjF )1020.1(m)(0.0300

    nC)nC)(6.0000.2(

    4

    14

    1 42

    0

    2

    23

    32

    0

    23 Nr

    v!

    !

    !T

    Therefore the two force components are:

    48

    v

    x

    v!vv!yF

    b) Thus, ))

    v!vv!! yx FFF and

    the angle is ,.)a tan( !! x FF below the xaxis

    21.69: a)

    !

    !

    222

    0

    2

    0

    2

    0 )1(

    1

    )1(

    1

    4

    1

    )(4

    1

    )(4

    1

    aaa

    qQ

    xa

    qQ

    xa

    qQ

    q

    .44

    1.))..21(...21(

    4

    13

    0

    2

    0

    2

    0

    xa

    qQ

    a

    x

    a

    qQ

    a

    x

    a

    x

    a

    qQ

    q

    !

    !} But this is the

    equation of a simple harmonic oscillator, so:

    am

    kqQ

    am

    qQ

    f

    am

    qQf !!!!

    b) If the charge was placed on they-axis there would be no restoring force ifq and Q

    had the same sign. It would move straight out from the origin along the y-axis, since the x-

    components of force would cancel.

  • 8/4/2019 Cap 21 Fisica Solucionario

    19/32

    21.70: Examining the forces: cosansin e !!!! mgTFFTF x

    So

    o

    d

    kq

    mgF !! But .t 31

    2

    23

    2 0

    22

    mg

    Lq

    mg

    Lkq

    Ld dd !!}

    21.71: a)

    b) Using the same force analysis as in problem 21.70, we find:

    gdq tan4

    2

    0

    2

    !and

    !

    rr!

    ! qqd )sm80.9)(kg015.0(25tan)25sin)2.1(2(4(1.2)sin25222

    0

    .1079.2 6v

    c) From Problem 21.70, .ta 22

    d

    kqmg !

    Lmg

    kq

    L

    d

    222

    2-629

    22

    2

    sin)smkg)(9.8015.0(m)6.0(4

    C)10)(2.79CNm1099.8(

    sin)2(tan

    2sin

    vv!!!

    Therefore

    2sin331.0tan ! . Numerical solution of this transcendental equation leads to

    ..

    !

    21.72: a) Free body diagram as in 21.71. Each charge still feels equal and opposite electric

    forces.

    b) .sinso.cos

    r

    q

    qT

    mgT e !!r!!r! (Note:

    )342.0m)sin20500.0(21 !r!r

  • 8/4/2019 Cap 21 Fisica Solucionario

    20/32

    c) From part (b), .C1071.3213

    21

    v!qq d) The charges on the spheres are made equal by connecting them with a wire, but we

    still have 22

    2

    041

    e N0453.0tan rQ

    mgF !!! where .

    221 qqQ

    ! But the separation2r is

    known: m)500.0(22!r m.500.030sin !r Hence: 2202 4

    21 rFQ eqq !! .C1012.1 6v!

    This equation, along with that from part (b), gives us two equations in1

    q and

    C1024.2. 6212v! qqq and .C1070.3 21321

    v!qq By elimination, substitution and

    after solving the resulting quadratic equation, we find: C1006.26

    1

    v!q and

    C1080.1 72v!q .

    21.73: a) g2.30)molgmol)(22.99100.0(mNaClmol100.0 Na !!

    g3.55)molgmol)(35.45100.0(mCl !!

    Also the number of ions is22

    A 1002.6mol)100.0( v!N so the charge is:

    C.9630C)1060.1)(1002.6(1922

    !vv!

    q The force between two such charges is:

    N.1009.2m)0200.0(

    )9630(

    4

    1

    4

    1 212

    2

    0

    2

    2

    0

    v!!!r

    q

    F

    b) .sm1089.5)kg10(3.55N)1009.2(223321 v!vv!! mFa

    c) With such a large force between them, it does not seem reasonable to think the

    sodium and chlorine ions could be separated in this way.

    21.74: a)

    v

    v

    !!v!

    2

    9

    2

    9

    32

    23

    32

    2

    13

    316

    3m)0.2(

    C105.4

    m)0.3(

    C)105.2(N100.4 kq

    r

    qkq

    r

    qkqF

    nC.2.3)CN(1262

    N100.4 63 !v!

    q

    b) The force acts on the middle charge to the right.

    c) The force equals zero if the two forces from the other charges cancel. Because of the

    magnitude and size of the charges, this can only occur to the left of the negative charge .2

    q

    Then:2

    2

    2

    12313

    )200.0()300.0( x

    kq

    x

    kqFF

    !

    ! where x is the distance from the origin.

    Solving forx we find: m.76.1!x The other value ofx was between the two chargesand is not allowed.

    21.75: a) ,6

    4

    1

    )2(

    )3(

    4

    12

    2

    02

    0 L

    q

    L

    qq

    F !! toward the lower the left charge. The other two

    forces are equal and opposite.

  • 8/4/2019 Cap 21 Fisica Solucionario

    21/32

    b) The upper left charge and lower right charge have equal magnitude forces at rightangles to each other, resulting in a total force of twice the force of one, directed toward thelower left charge. So, all the forces sum to:

    .2

    323

    4)2(

    )3(2)3(

    4

    12

    0

    2

    22

    0

    !

    !

    L

    q

    L

    qq

    L

    qq

    F

    21.76: a) .2

    )()(4

    1)(

    222

    0

    !

    y

    q

    ay

    q

    ay

    q

    pE

    b) ).2)1()1((

    4

    1)( 22

    2

    0

    ! yaya

    y

    q

    p

    Using the binomial expansion:

    .4

    12

    321

    321

    4

    1)(

    4

    2

    0

    2

    2

    2

    2

    2

    0 y

    qa

    y

    a

    y

    a

    y

    a

    y

    a

    y

    q

    p

    !

    }

    Note that a point charge drops off like2

    1

    yand a dipole like .

    13y

    21.77: a) The field is all in the x -direction (the rest cancels). From the charges:

    .)(4

    1

    4

    1

    4

    12322

    02222

    0

    22

    0 xa

    qx

    xa

    x

    xa

    q

    E

    xa

    q

    E x

    !

    !

    !

    (Each q contributes this). From the q

    !

    !! xa

    x

    q

    x

    q

    xa

    qx

    E

    x

    q

    E

    x

    b) ..for,3

    4

    11

    2

    31

    2

    4

    14

    2

    0

    2

    2

    2

    0

    total axx

    qa

    x

    a

    x

    q

    E ""!

    }

  • 8/4/2019 Cap 21 Fisica Solucionario

    22/32

    Note that a point charge drops off like x

    and a dipole like x

    .

    21.78: a) 20.0 g carbon .og0.2

    0.20!

    gmol carbon 0.10)67.1(6 ! mol electrons

    .100. 6)1060.1()0.10( 61

    A

    !

    !

    N This much charge is placed at the earths

    poles (negative at north, positive at south), leading to a force:

    .1013.)m10276.1(

    C)10963.0(

    4

    1

    )2(4

    1 727

    26

    0

    2

    earth

    2

    0

    vv

    v

    R

    F

    b) A positive charge at the equator of the same magnitude as above will feel a forcein the south-to-north direction, perpendicular to the earths surface:

    m)

    )

    sin)

    earth

    v

    v

    v

    F

    R

    q

    F Q

    21.79: a) With the mass of the book about 1.0 kg, most of which is protons and neutrons,we find: #protons = .100.3kg)1067.1(kg)0.1(

    2627

    21

    v!v

    Thus the charge difference

    present if the electrons charge was %999.99 of the protons is

    .0)106.1)(00001.0)(100.( 1926 !vv!( q

    b) raThr pu s v38))48) 3

    v

    ( krqkF

    sm103.8)kg1()N103.8(21313

    v!v!! mF

    c) Thus even the slightest charge imbalance in matter would lead to explosive

    repulsion!

    21.80:(a)

  • 8/4/2019 Cap 21 Fisica Solucionario

    23/32

    22

    0

    2

    22

    22

    0

    2

    22

    0

    2

    2

    2

    0

    2

    2

    0

    net

    )()(

    4

    4)()(

    )()(

    4

    )(

    1

    )(

    1

    4

    )(4

    1

    )(4

    1charge)central(on

    xbxb

    bx

    q

    xbxb

    xbxb

    q

    xbxb

    q

    xb

    q

    xb

    q

    F

    !

    !

    !

    !

    For ,bx this expression becomes

    .tooppositeisDirection3

    0

    2

    22

    0

    2

    net xxb

    q

    bb

    bx

    qF !

    (b) 22

    30

    2

    :dt

    xd

    b

    qmxmaF !!

    TI

    3

    0

    2

    3

    0

    2

    3

    0

    2

    2

    2

    2

    12bm

    q

    ff

    bm

    q

    xbm

    q

    dt

    xd

    !p!!

    !

    (c) kg)1066.1(12amu12m,100.4,2710 v!!v!! mbeq

    z1028.4)m100.4)(NmC1085.8()kg1066.1(12

    )C106.1(

    2

    1 12310221227

    219

    v!vvv

    v!

    f

    21.81: a) !vv!!! 3334333

    34 )m1000.1)()(mkg109.8()( rVm

    kg10728.3 5v

    mol10867.5)molkg10546.63)(kg10728.3( 435 v!vv!! Mmn

    atoms105.3 20A v!! nNN (b) protonsandelectrons10015.1)105.3)(29(

    2220

    e v!v!N

    C6.1)10015.1)(C10602.1)(10100.0()99900.0( 22192eenet !vvv!!

    eNeNq

    N103.2)m00.1(

    )C6.1( 102

    2

    2

    2

    v!!! kr

    qkF

    21.82: First, the mass of the drop:

    kg.1041.13

    )m100.15(4)mkg1000( 11

    363

    v!

    v!!

    m

    Next, the time of flight: :onacceleratitheands00100.02002.0 !!! vDt

    .sm600)s001.0(

    )m1000.3(22

    2

    1 22

    4

    2

    2 !v

    !!!

    t

    daatd

    So:

    C.1006.1CN1000.8

    )sm600)(kg1041.1( 134

    211

    v!v

    v!!!! EmaqmqEmFa

  • 8/4/2019 Cap 21 Fisica Solucionario

    24/32

    21.83: eEFy ! 0!xF

    pp m

    eE

    m

    Fa

    y

    y!! 0!

    xa

    a) ymeEvyavvp

    yyy (!(!2

    sin2 220202 0when ax !!( yvhy

    eE

    vh

    p

    2

    sin 22omax !

    b) 202

    1tatvy yy !(

    ig ig

    ig

    i

    whe

    tta

    tt

    y

    !

    !(!

    ya

    vt sin2,so

    orig !

    or

    eE

    mvtvd

    eE

    mvt

    x sinos2

    sin2

    p

    2

    0

    orig0

    p0

    orig

    !!

    !

    c)

    d) m42.0)N500)(106.1(2

    30sin)kg1067.1()sm104(19

    22725

    m x !v

    rvv!

    h

    m89.2)CN500)C106.1

    30s n30cos)kg1067.1)sm104219

    2725

    !v

    rrvv

    ! d

    21.84: a) !

    !!! 22

    2

    2

    2

    1

    1

    0

    2

    2

    2

    0

    2

    1

    1

    0 4

    1

    4

    1

    4

    1CN50 q

    r

    q

    r

    q

    r

    q

    r

    q

  • 8/4/2019 Cap 21 Fisica Solucionario

    25/32

    C.108)m6.0(

    )C1000.4(CN0.504)m2.1(4

    9

    2

    9

    0

    2

    22

    1

    10

    2

    2

    v!

    v!

    q

    r

    qEr

    b) !

    !!! 22

    2

    2

    2

    1

    1

    0

    2

    2

    2

    0

    2

    1

    1

    04

    1

    4

    1

    4

    1CN50 q

    r

    q

    r

    q

    r

    q

    r

    q

    E

    C.100.24)m6.0(

    )C1000.4()0.50(4)m2.1(

    50 92

    9

    0

    2

    22

    1

    12

    2

    v!

    v!

    q

    r

    q

    kr

    21.85:222

    )m00.5()m00.8(

    nC)0.12(

    )m00.3(

    )nC0.16(CN0.12

    kqkkE

    !!

    nC.1.73C1031.7m00.64

    C1020.1

    m0.9

    C1060.112m0.25 8

    2

    8

    2

    82 !v!

    v

    v!

    kq

    21.86: a) On the x-axis: !!!a

    xxxraa

    Qdx

    Era

    dq

    dE0

    2

    0

    2

    0)(4

    1)(4

    1

    0nd.11

    4

    1

    0

    !

    yE

    rara

    Q

    .

    b) .11

    4

    111

    4

    1then,I

    00

    iEF

    !!

    !!

    xaxa

    qQ

    q

    xaxa

    Q

    Exra

    TT

    c) }}!!"" 2

    1 )11()1)1((,Forx

    kqQxa

    ax

    kqQxa

    ax

    kqQFax

    pointalikelooksondistributiCharge.),forthatNote(.4

    12

    0 xaxraxr

    qQ

    }!""

    from far away.

    21.87: a) !

    !

    !

    ! yx dEyxa

    dykQxdE

    yxa

    dykQ

    yx

    dqkdE and

    (with

    )()(23222222

    .)(

    2322yxa

    dyKQy

    Thus:

    TI!TI!!a

    xaxx

    Q

    x

    a

    axa

    Qx

    yx

    dy

    a

    Qx

    E

    0

    2122

    0

    22122

    0

    2322

    0)(4

    1

    )(

    1

    4

    1

    )(4

    1

    !

    ! 212200

    2322

    0)(

    11

    4

    1

    )(4

    1

    axxa

    Q

    yx

    ydy

    a

    Q

    E

    a

    y

    b) (a).ingivenareandhereand yxyyxx EEqEFqEF !!

    c) .24

    1

    24

    1))1(1(

    4

    1,or

    3

    0

    2

    2

    0

    2122

    0 x

    qQa

    x

    a

    ax

    qQ

    xa

    ax

    qQ

    Fax y !}!""

  • 8/4/2019 Cap 21 Fisica Solucionario

    26/32

    Looks dipole-like iny-direction .4

    14

    1

    2

    0

    21

    2

    2

    2

    0 x

    qQ

    x

    a

    x

    qQ

    Fx }

    !

    Looks point-like along x-direction

    21.88: a) iE 4

    1),9.22(.From22

    0 axxQ

    Eq

    !T

    .)CN1029.1(

    )m025.0()m105.2()m105.2(

    )1000.9(

    4

    1 62233

    9

    0

    iiE v!vv

    v!

    C

    T

    (b) The electric field is less than that at the same distance from an infinite line of

    charge ).CN1030.12

    2

    4

    12

    4

    1( 6

    00

    v!

    !!gpax

    Q

    xEa This is because in the

    approximation, the terms left off were negative. !

    }

    22

    0

    21

    02

    1212

    12

    2

    ax

    xxa

    x

    gLine

    E (Higher order terms).

    c) For a 1 difference, we need the next highest term in the expansion that was left off

    to be less than 0.01:

    cm.0.352(0.01)m025.0)01.0(201.02 2

    2

    ! xaxa

    x

    21.89: (a) From Eq. (22.9), iE 4

    122

    0 axx

    Q

    !T

    .C)N7858()m025.0(m)(0.100m)(0.100

    C)100.9(

    4

    122

    9

    0

    iE !

    v!

    T

    b) The electric field is less than that at the same distance from a point charge (8100

    point2

    2

    2

    02

    14

    1ESinceC).N E

    x

    a

    x

    Q

    x

    !

    !gp (Higher order terms).

    c) For a 1 difference, we need the next highest term in the expansion that was left

    off to be less than 0.01:

    .m177.002.01025.0))01.0(2(101.02 2

    2

    }!}} xaxx

    a

  • 8/4/2019 Cap 21 Fisica Solucionario

    27/32

    21.90: (a) On the axis,

    !

    !

    21

    2

    2

    0

    221

    2

    2

    0

    1)m0020.0(

    )m025.0(1

    2

    )m025.0(Cp00.411

    2

    x

    R

    E

    ! thein,CN106E x-direction.

    b) The electric field is less than that of an infinite sheet .CN1152 0

    !!g

    E

    c) Finite disk electric field can be expanded using the binomial theorem since the

    expansion terms are small:

    }3

    3

    02

    12 R

    x

    R

    x

    E So the difference between the

    infinite sheet and finite disk goes like .R

    xThus:

    .1616.05.24.0

    cm)40.0(and80.082.50.2cm)20.0(

    !!}

    !(!!}!( xExE

    21.91: (a) As in 22.72:

    !

    21

    2

    2

    0

    112 x

    E

    E

    !

    21

    2

    2

    0

    2

    1)m200.0(

    m)025.0(1

    2

    m)(0.025pC00.4

    ! in theCN89.0 x-direction.

    b) )]8321(1[2

    , 4422

    0

    !"" xRxR

    ERx

    .4422

    2

    0

    2

    0

    2

    2

    2

    0 x

    Q

    x

    R

    x

    R

    !!}

    c) The electric field of (a) is less than that of the point charge (0.90 )CN since the

    correction term that was omitted was negative.

    d) From above m1.0For.101.00.89

    0.89)(0.9m2.0 !!!

    ! xx

    CN6.3

    CN43.3

    point

    disk

    !

    !

    E

    E

  • 8/4/2019 Cap 21 Fisica Solucionario

    28/32

    so .5047.06.3

    )43.36.3(}!

    21.92: a)

    !!!

    a aa

    a axdxfdxxfdxxfdxxfxfxf

    0 0

    0

    )()()()()(:)()(

    !!

    a

    a a

    a

    a a

    dxxf

    dxxfydyfdxxfyxdxxf

    0

    0 0 0

    .)(2

    )()()()(:ithreplaceNo.)(

    b)

    !!!a

    a a

    a a

    xdxgdxxgdxxgdxxgxgxg0

    0 0))()(()()()(:)()(

    !!a

    a

    a aa

    dxxgydygdxxgyxdxxg0 00

    .0)()()()(:withreplaceNow.)(

    c) The integrand in yE for Example 21.11 is odd, so yE =0.

    21.93: a) The y-components of the electric field cancel, and the x-components from bothcharges, as given in problem 21.87 is:

    .)(

    112

    4

    1

    )(

    112

    4

    12122

    0

    2122

    0

    iF

    !

    !ayya

    Qq

    ayya

    Q

    Ex

    T

    If .4

    1))21(1(2

    4

    1,

    3

    0

    22

    0 y

    Qqa

    ya

    ay

    Qq

    ay !

    }"" iFT

    b) If the point charge is now on thex-axis the two charged parts of the rods provide

    different forces, though still along the x-axis (see problem 21.86).

    iEFiEF 11

    4

    1and

    11

    4

    1

    00

    !!

    !! axxa

    Qq

    q

    xaxa

    Qq

    q

    TTTT

    So,

    iFFF 121

    4

    1

    0

    !! axxaxa

    Qq

    TTT

    For .2

    4

    1...12...14

    1,

    3

    0

    2

    2

    2

    2

    0

    iiFx

    Qqa

    x

    a

    x

    a

    x

    a

    x

    a

    ax

    Qq

    ax !

    }""

    T

    21.94: The electric field in the x-direction cancels the left and right halves of the

    semicircle. The remaining y-component points in the negative y-direction. The charge per

    unit length of the semicircle is:

  • 8/4/2019 Cap 21 Fisica Solucionario

    29/32

    .sin

    sinbutand2

    a

    dkdEdE

    a

    dk

    a

    dlkdE

    a

    Qy

    !!!!!

    So, !!U!!

    o,o

    ya

    kQ

    a

    k

    a

    kd

    a

    kE

    21.95: By symmetry, yx EE ! For yE compared to problem 21.94, the integral over the

    angle is halved but the charge density doublesgiving the same result. Thus,

    .22

    2a

    kQ

    a

    kEE yx !!!

    21.96:

    mgTmgTF

    x

    cos

    cos0

    !

    !!

    !

    !

    !

    mg

    q

    mg

    q

    q

    mg

    q

    T

    q

    TFy

    0

    00

    00

    2arctan

    2tan

    sin2cos

    sin22sin0

    21.97: a) 29)sm8.9(0

    CN104.1

    g1010

    2

    5

    !v

    !!!E

    q

    mmgqE

    b) 1429 .e e

    rb11.1

    e e

    1.1.

    l

    rb12..

    112

    l1. 1

    192

    v

    vv

    v ee

    21.98: a) yx EE ! and

    ere,4

    122

    2

    2

    20

    c

    ar

    e, ireoflen

    t

    aQax

    xx

    Q

    EE

    !!

    .,

    a

    QE

    a

    QE

    ax yx

  • 8/4/2019 Cap 21 Fisica Solucionario

    30/32

    b) If all edges of the square had equal charge, the electric fields would cancel by

    symmetry at the center of the square.

    21.99: a)0

    2

    0

    2

    0

    2

    0

    3

    0

    2

    0

    1

    2

    mC0200.0

    2

    mC0100.0

    2

    mC0200.0

    2

    2

    2

    )(

    PE !!

    v!! in theC,N1065.52

    mC0100.0)(

    8

    0

    2

    PE x-direction.

    b)0

    2

    0

    2

    0

    2

    0

    3

    0

    2

    0

    1

    2

    mC0200.0

    2

    mC0100.0

    2

    mC0200.0

    2

    2

    2

    )(

    RE !!

    v!! in theC,N1069.12

    mC0300.0)(

    9

    0

    2

    RE x-direction.

    c)0

    2

    0

    2

    0

    2

    0

    3

    0

    2

    0

    1

    2

    mC0200.0

    2

    mC0100.0

    2

    mC0200.0

    2

    2

    2

    )(

    SE !!

    v!! in theC,N1082.22

    mC0500.0)( 9

    0

    2

    SE x-direction.

    d)0

    2

    0

    2

    0

    2

    0

    3

    0

    2

    0

    1

    2

    mC0200.0

    2

    mC0100.0

    2

    mC0200.0

    2

    2

    2

    )(

    TE !!

    v!! in theC,N1065.52

    mC0100.0)(

    8

    0

    2

    SE x-direction.

    21.100: m.N1013.12

    mC1000.2

    2

    7

    0

    24

    0

    32

    1IatIon v!

    v!

    !!

    A

    qE

    A

    F

    mN1026.22

    mC1000.4

    2

    7

    0

    24

    0

    312IIatIIon v!v

    !

    !!

    A

    qE

    A

    F

    mN1039.32

    mC1000.6

    2

    7

    0

    24

    0

    21

    3IIIatIIIon v!

    v!

    !!

    A

    qE

    A

    F

    (Note that + means toward the right, and is toward the left.)

    21.101: By inspection the fields in the different regions are as shown below:

    ).(2

    )(2

    ),(2

    )(2

    ),(2

    0

    00

    00

    kiE

    kiki

    kiki

    z

    z

    x

    x

    E

    E

    E

    E

    IVIII

    III

    !@

    !

    !

    !

    !

    T

  • 8/4/2019 Cap 21 Fisica Solucionario

    31/32

    21.102: a) RRAQ )(2

    2

    2

    2

    b) Recall the electric field of a disk, Eq. (21.11): ? A o,.1)(112

    2

    0

    xR

    E

    ? A ? A

    .)()(

    )()()()(

    i

    iE

    x

    xxRxR

    xE

    x

    xxRxR

    x

    v!

    !T

    c) Note that

    }! ...

    2

    )1))11)1

    2

    1

    1

    212

    1

    1

    2

    1

    Rx

    R

    xRx

    R

    xxR

    ,11

    2

    2)(

    2210210

    ii

    x

    x

    x

    x

    x

    x

    xE

    !

    !

    Xand sufficiently close means that

    Rx

    d) )(

    !!!

    !!

    RRm

    q

    fmx

    RR

    qxq

    F

    21.103: a) The four possible force diagrams are:

    Only the last picture can result in an electric field in the x-direction.b) .and.4. 3 "!! qqq

    c)22

    2

    0

    12

    1

    0

    sinm)0300.0(4

    1sin

    )m0400.0(4

    10

    q

    q

    Ey !!

    C.43.064

    27

    54

    53

    16

    9

    sin

    sin

    16

    911

    2

    112 qq

    qq !!!!

  • 8/4/2019 Cap 21 Fisica Solucionario

    32/32

    d) .565

    3

    9.5

    4

    16.4

    1 1

    333 !

    !!qq

    qEqF

    x

    21.104: (a) The four possible diagrams are:

    The first diagram is the only one in which the electric field must point in the negativey-direction.

    b) .0nd00.3 21 ! qq

    c)12

    5

    )050.0()120.0(13

    12

    )120.0(13

    5

    )050.0(0

    2

    1

    2

    2

    2

    2

    2

    1 kqkqkqkqEx

    !!!

    !!!13

    5

    12

    5

    13

    12

    m)05.0(13

    5

    m)120.0(13

    12

    m)050.0( 21

    2

    2

    2

    1 kqkqkqEEy

    C.N1017.17

    v!! yEE

    21.105: a) For a rod in general of length

    !rLrL

    kQEL

    11, and here .

    2

    axr !

    So, .ro!" # $ %

    !

    !axLaxL

    kQ

    axLaxL

    kQE

    b)

    v!!!&!2

    2

    2

    22

    22aL

    a

    aL

    aL

    k'dx

    L

    E'dqE

    (

    Edqd(

    dxaxLax

    22

    1

    2

    1

    )

    )

    )

    )

    )

    )

    )][l ()](l[ aLaaL

    a axLxaL

    k0F !1

    .)(

    )(2

    2

    2

    2

    2

    3

    3

    4

    3

    3

    33

    4

    5

    Laa

    La

    L

    kQ

    aL

    aL

    a

    aLa

    L

    kQF

    c) For ))21ln()2(1n12()21(

    )1(n1:

    2

    2

    2

    22

    2

    2

    aLaLL

    kQ

    aLa

    aLa

    L

    kQFLa !

    !""

    6

    6

    6

    6

    6

    6

    6

    6

    a

    kQF

    a

    L

    a

    L

    a

    L

    a

    L

    L

    kQF }

    }